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I need to find partial fractions for $$\frac{1}{x^4 + 1}$$ so it can be integrated, but I'm completely stuck.

Davide Morgante
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user486454
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  • There a tricky method that is easy to follow but may be hard to think of in the first place, and it requires no knowledge of complex numbers. There's also a straightforward method that requires complex numbers, including in particular de Moivre's formula and knowing what happens with complex-conjugate pairs. I've posted the straightforward method; several others have posted the more elementary by tricky method. – Michael Hardy Sep 02 '18 at 18:51

4 Answers4

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Hint: $$x^4+1=(x^2+1)^2-2x^2=(x^2-\sqrt 2x+1)(x^2+\sqrt 2x+1).$$

Bernard
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Hint:

Don't really need to factorize

write numerator as $$\dfrac{x^2+1-(x^2-1)}2$$

$$\dfrac{d\left(x\pm\dfrac1x\right)}{dx}=\text{?}$$

$$\dfrac2{x^4+1}=\cdots=$$ $$\dfrac{1+\dfrac1{x^2}}{x^2+\dfrac1{x^2}}$$ $$-\dfrac{1-\dfrac1{x^2}}{x^2+\dfrac1{x^2}}$$

Michael Hardy
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lab bhattacharjee
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Hint: Write $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2$$ and use the binomial formula.

Dr. Sonnhard Graubner
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    The binomial formula? The thing to use here is factoring a difference of two squares. I wouldn't call that the binomial formula. – Michael Hardy Sep 02 '18 at 19:17
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Several people have pointed out that \begin{align} x^4+1 & = (x^4 +2x^2+1) - (2x^2) \\[10pt] & = (x^2+1)^2 - (x\sqrt 2)^2 \\[10pt] & = (x^2+1-x\sqrt 2\,)(x^2+1 -x\sqrt 2\,). \\[10pt] \end{align} So I will point out a more straightforward approach, albeit one that requires some awareness of complex numbers. \begin{align} x^4 + 1 & = 0 \\ x^4 & = -1 \\ x^2 & = \pm i \\[8pt] \text{If } x^2 & = i = \cos90^\circ+i\sin90^\circ \\ \text{then } x & = \pm(\cos45^\circ+i\sin45^\circ) = \pm\left( \frac{\sqrt2} 2 + i \frac{\sqrt2} 2 \right) \\[12pt] \text{and similarly if } x^2 & = -i, \\ \text{then } x & = \pm(\cos45^\circ - i \sin45^\circ) = \pm\left( \frac{\sqrt2} 2 - i \frac{\sqrt2} 2 \right). \end{align}

Next observe what happens with complex-cojugate pairs: the imaginary parts cancel when you multiply them: $$ \left( x - \frac{\sqrt2}2 + i\frac{\sqrt2}2 \right) \left( x - \frac{\sqrt2}2 -i\frac{\sqrt2} 2 \right) = x^2 - x\sqrt2+1 $$ and similarly for the other pair.

(That's how I did this problem the first time I ever saw it.)

Michael Hardy
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